I am working on a function which I want to find a nice lower bound for it. The function is $$ \left(\lceil \frac{x}{2}\rceil-\left|x-2\lceil\frac{x}{2}\rceil\right|\right)^2, \qquad \text{for }x\geq 0 $$ By plotting graph, it turns out that this function can be bounded below by $\left(\frac{x-1}{2}\right)^2$. However, I have trouble in working this bound analytically. I have been trying to use the inequality $\lceil x\rceil-1<x\leq \lceil x\rceil$, but it is not getting to the result.
Can anyone help me out?
You function is basically
$$f(x)=(x-n)^2\ \ \text{if}\ \ x\in(2n-2,2n]\ \ \text{for}\ \ n\in\mathbb N^+.$$
To see this, use the facts $\lceil x/2\rceil=n$ for $x\in(2n-2,2n]$ and $|x-2\lceil x/2\rceil|=2\lceil x/2\rceil-x$.
To find a lower bound, we need to find the minimum of $f(x)$ on each interval $x\in(2n-2,2n]$, i.e. $\inf_{x\in(2n-2,2n]}f(x)$. Since $f$ is piecewise quadratic, we simply get
If $n=1$, then $\inf_{x\in(0,2]}f(x)=\inf_{x\in(0,2]}(x-1)^2=0$ when $x=1$,
If $n\geq2$, then $\inf_{x\in(2n-2,2n]}f(x)=\inf_{x\in(2n-2,2n]}(x-n)^2=(n-2)^2$ when $x=2n-2$.
So the minimum points are $(1,0),(2,0),(4,1),(6,4),...,(2n-2,(n-2)^2),...,$ based on these we can easily find a lower bound $L$ by
$$L(x)=\left\{\begin{matrix}(x-1)^2,&0\leq x\leq1\\0,&1<x\leq2\\\frac14(x-2)^2,&x>2\end{matrix}\right..$$
Here is the Desmos plot for this bound.
If you don't want a piecewise bound, simply use $L(x)=\frac{(x-2)^2-2}4$.